# Equivalent Representations of Clifford Algebra

I'm reviewing David Tong's excellent QFT lecture notes here and am a little confused by something he writes on page 94.

We've considered the standard chiral representation of the Clifford Algebra, and he is now generalising to a different representation. He writes that this will involve

$$\gamma^{\mu}\rightarrow U\gamma^{\mu}U^{-1} \ \textrm{and} \ \psi \rightarrow U\psi$$

What does the second transformation mean? I don't see how it has anything to do with a representation of the Clifford algebra (which is an assignment of a matrix to every element of the algebra, as far as I know). Does he mean that in the resulting projective representation of the Lorentz group we should transform

$$\psi\rightarrow S[\Lambda]U\psi$$

or am I barking up the wrong tree?

This is all very reminiscent of changing pictures in QM, but I've never talked about representation theory in that context! Is there a rigorous link?

Many thanks!

• Dear Edward, it's extremely hard to see some beef in your question. Correct me if I am wrong but what you seem to be puzzled by is the undergraduate freshmen's first-semester linear algebra. $\psi\to U\psi$ is just a change of basis on the space of vectors $\psi$. After all, the transformation rule for the operators on the same space $\gamma^\mu\to U\gamma^\mu U^{-1}$ is derived from $\psi\to U\psi$, so it is very hard how you can possibly understand the rule for the gammas but not for the psis. And why do you talk about projective reps, much more complicated concepts than plain basis change? – Luboš Motl Dec 24 '12 at 12:02
• Maybe the problem is that you are confusing the matrix $U$ in your transformation formulae with the action of a Lorentz or another element $\Lambda$. They are completely different transformations. The $U$ in the transformation laws you indicated is just a change of basis, a change of representation, and this $U$ doesn't depend on any $\Lambda$ in the Lorentz group or something like that. On the contrary, it is some fixed matrix, like a permutation matrix with $\pm 1$ or $\pm i$ entries, or discrete Fourier transform etc., and is used to change the basis/representatino for all $\Lambda$. – Luboš Motl Dec 24 '12 at 12:11
• Ah okay - I see my problem. I naturally think of a representation as a linear action of a group on a vector space, which automatically identifies all equivalent representations. Changing between equivalent representations is merely effecting a change of basis on the vector space. In the new basis, of course, $\psi$ becomes $U\psi$ and $\gamma^{\mu}$ is written $U\gamma^{\mu}U^{-1}$, by standard linear algebra. I tripped up because I was trying to think about genuinely different representations on the same space, which wouldn't just boil down to a change of basis in general. – Edward Hughes Dec 24 '12 at 12:16
• And thanks very much for your help - sometimes I overthink something simple and need the easy argument pointed out! – Edward Hughes Dec 24 '12 at 12:17
• Right, good for you, Edward, and Merry Christmas. Incidentally, changing momentum/position representation in QM is also an example of a change of basis, and these equivalent reps may also be thought of as the "same thing" following your abstract, basis-independent perspective. – Luboš Motl Dec 24 '12 at 12:27

## 1 Answer

I was really confused by something like this too (it was a statement that $Y_{lm}$ was 'a representation' of a group of rotations). The problem is that in physics textbooks the distiction between a group and an acton of this group is usually not stressed enough.

In your case we are considering two representations (lets call them $A$ and $B$). For an element $g_\Lambda$ of the Lorentz group the corresponding linear operators would act on a vector space of Dirac spinors:

$$A(g_\Lambda)\cdot\psi = S[\Lambda]^\alpha_\beta \psi^\beta,\quad B(g_\Lambda)\cdot\psi = S'[\Lambda]^\alpha_\beta \psi^\beta$$ (I'm dropping coordinate dependence, and I won't write indices any more.)

Now we are saying that the two representations $A$ and $B$ are (unitary) equivalent if there is a unitary transformation $U$ which allows you to 'compensate' for the switch of those representations. So let us take two spinors $\psi$ and $S[\Lambda]\psi$ and transform them like: $$\psi\to U\psi,\quad S[\Lambda]\psi\to US[\Lambda]\psi$$ But also we have: $$A\to B \quad\Rightarrow\quad S[\Lambda]\to S'[\Lambda]$$ So, for consistency, we must have: $$US[\Lambda]\psi = S'[\Lambda]U\psi \quad\Rightarrow\quad S'[\Lambda] = US[\Lambda]U^{-1}$$ Finally, gathering stuff up: $$\psi\to U\psi$$ $$S[\Lambda]\to US[\Lambda]U^{-1}$$